Insert Interval
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1: Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2: Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].
This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
思路： 因为区间按 start 升序，且无重叠。所以插入区间和每一个元素分三种情况考虑。在左边，在右边（此两种情况直接拿区间出来）或者交叉（则更新插入区间范围）。 利用变量 out 判断新的区间是否已经放入。
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
class Solution {
public:
vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
vector<Interval> vec;
bool out = true;
for(size_t i = 0; i < intervals.size(); ++i) {
if(intervals[i].end < newInterval.start) {
vec.push_back(intervals[i]);
} else if(intervals[i].start > newInterval.end) {
if(out) { vec.push_back(newInterval); out = false;}
vec.push_back(intervals[i]);
} else {
newInterval.start = min(newInterval.start, intervals[i].start);
newInterval.end = max(newInterval.end, intervals[i].end);
}
}
if(out)
vec.push_back(newInterval);
return vec;
}
};
Merge Intervals
Given a collection of intervals, merge all overlapping intervals.
For example, Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18].
思路： 先按 start 排序。然后，判断当前区间和前一区间是否重叠。若没重叠，则放入；若重叠，则更新前一区间 end, 舍弃当前区间。
/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
bool cmp(Interval a, Interval b) {
return a.start < b.start;
}
class Solution {
public:
vector<Interval> merge(vector<Interval> &intervals) {
sort(intervals.begin(), intervals.end(), cmp);
vector<Interval> vec;
for(size_t i = 0; i < intervals.size(); ++i) {
if(vec.empty()) vec.push_back(intervals[i]);
else if(intervals[i].start <= vec.back().end)
vec.back().end = max(intervals[i].end, vec.back().end);
else vec.push_back(intervals[i]);
}
return vec;
}
};
